RE: Capacitor Sizing for Power

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Loudner, Godfrey by way of Terry Fritz <twftesla-at-qwest-dot-net>" <gloudner-at-SINTE.EDU>

Hi Matt 

My previous post uses a wrong number. You wrote that 15KV*240mA = 2700 W,
but 15KV*240mA = 3600 W. Your main equation amount to VI(1sec.) =
120(0.5)CV^2. This gives C = I(1sec)/(60V) = 0.240amp(1sec)/(60*15,000volt)
= 0.26667uF. Now I will redo my previous post.

All your first equation says is that 120 caps with a capacitance of
0.26667uF placed in parallel will be able to store 3600 joules of energy at
a potential of 15,000 volts. This is already enough to lock in the charging
time of a cap, say 
5RC = 5(V/I)*I(1sec)/(60V) = (1/12)sec. You are asking for all 120 caps to
charge and discharge in 1sec. This is the same as asking a single cap to
charge and discharge in (1/120)sec, or a charging time of (1/240)sec. Now
(1/12)sec is 2000% of (1/240)sec. So you want to place a 0.26667uF cap in a
tank circuit that requires (1/12)sec to charge, but you are asking it to
forget all about that and charge in (1/240)sec.

Notice that your main equation will always give a charging time of (1/12)sec
for any V and I. 

I cannot fault Matt for his arithmetic, because in my previous post I wrote
5(15,000volt/0.240amp)(0.0000002F) = 0.0036sec. This is in error because I
multiplied 15,000 with 0.240 when I should have divided 15,000 by 0.240.
Correctly
5(15,000volt/0.240amp)(0.0000002F) = 0.0625sec = (1/16)sec. Its been an
arithmetically bad night. :-(

Godfrey Loudner 



> -----Original Message-----
> From:	Tesla list [SMTP:tesla-at-pupman-dot-com]
> Sent:	Wednesday, July 18, 2001 5:54 PM
> To:	tesla-at-pupman-dot-com
> Subject:	Capacitor Sizing for Power
> 
> Original poster: "by way of Terry Fritz <twftesla-at-qwest-dot-net>"
> <Mddeming-at-aol-dot-com>
> 
> Hi All! 
>          
>         I just started working up the requirements for my 15/240 NST farm 
> coil and decided to try the P=rate*0.5*C*V^2 method. using a SRSG -at-120
> BPS. I 
> have read that this is a "good ballpark figure" for minimum cap size
> needed 
> to handle the power. 
> Now 15KV*240mA= 2700 W Substituting gives us 
> 2700=120*0.5*15000^2*C  or 
> 2700=60*15*15*1E6*C which reduces to   
> 3/15*1E-6=C  or 
> 0.20uF=C = 200nF 
> 
> Using the resonance relationship to calculate the minimum capacitance: 
> XL=2*pi*f*L=Xc=1/(2*pi*f*C) 
> if R<<XL then Z=E/I ~XL 
> XL=15000/.240=62500=377*L or L=166.8Hy 
> Xc=1/(2pi*f*C) 
> 62500=1/377C or 2.35625*1e7=1/C 
> 4.244*1e-8=C or, 
> C=42.44nF 
> A difference of 470% is not close in my ballpark. Even doubling the value
> for 
> LTR gives an error >230% 
> Am I missing something obvious or am I just brain dead this week? 
> 
> Matt D.
>